Is it true that cartesian product of any two circle 1 define a torus?

Question

I mean , I understand that torus is cartesian product of two circle 1 , shape like donut but I think we need two circle with different radious.So that it should not be true that any product of two circle is a torus.

Answer 1

I think there are two possible notions of a torus with radii $r,s$ which you might be confusing.

1) If $r\le s$, consider the set $T(r,s) \subset \mathbb{R}^3$, which is obtained by rotating the circle $\{(x,0,z) ~ \vert ~ (x-s)^2 + z^2 = r^2\}$ around the $z-$axis.

2) The Riemannian manifold $\tilde T(r,s) = S^1_{r} \times S^1_{s}$. Here $S^1_t$ denotes a circe with radius $t$ viewed as a Riemannian manifold and $\tilde T(r,s)$ carries the product Riemannian metric.

Let's study how these two sets are related. First note that $T(r,s)$ is a submanifold of $\mathbb{R}^3$ if and only if $r<s$, otherwise it has a self intersection at the origin. If $r<s$ it also inherits the structure of a Riemannian manifold from the ambient space. This raises the following questions:

a) Are $T(r,s)$ and $\tilde T(r',s')$ homeomorphic? (This means that one can continuously be deformed into the other, without bothering about preserving distances etc.)

If $r<s$ and $r',s'$ arbitrary, then the answer is yes, both are topologically a torus. If $r=s$, then $T(r,s)$ is not a manifold and thus there is no pair $r',s'$ such that the two objects are homeomorphic.

b) Are $T(r,s)$ and $\tilde T(r',s')$ diffeomorphic? (This only makes sense for $r>s$ and means that the deformation can be performed smoothly.)

If $r<s$ and $r',s'$ arbitrary, then the answer is yes, this is as easy to see as a) because the obvious homeomorphism is already smooth.

c) Are $T(r,s)$ and $\tilde T(r',s')$ isometric? (This means that there exists a smooth deformation from one into the other that preserves it's geometric properties like distances, areas and curvature.)

In general no, you might say, but what if $r=r'$ and $s=s'$? Not even then, but to see this it's better to know a little more geometry. The idea is that $T(r,s)$ has an intrinsic curvature. On the other hand circles $S^1_t$ can locally be flattened, which means that a product $\tilde T(r,s) = S^1_{r} \times S^1_{s}$ is also flat, i.e. has no intrinsic curvature. Since isometries preserve curvature, the two objects cannot be isometric.